Certain chemical reactions can protect people from serious injury or death. The following chemical r — Physical Chemistry Chemistry Question
Lifesaving chemistry of the airbag
Certain chemical reactions can protect people from serious injury or death. The following chemical reactions used to be utilized to rapidly produce large amounts of nitrogen gas inside an automobile airbag:
2 NaN3 → 2 Na + 3 N2(g) (1)
10 Na + 2 KNO3 → K2O + 5 Na2O + N2(g) (2)
K2O + Na2O + SiO2 → alkaline silicate (“glass") (3)
Write the Lewis structure for the azide anion and nitrogen molecule.
Model Answer
[VISUAL]
How many grams of sodium azide are needed to generate enough nitrogen to fill an airbag with a volume of 15,0 dm3 at a temperature of 50 °C and a pressure of 126,6 kPa?
Model Answer
n(N2) = (126.6 kPa * 15,0 dm3) / (8.314 J mol-1 K-1 * 323 K) = 0.707 mol
Mass of sodium azide needed to generate 0.707 moles of nitrogen:
n(NaN3) = 2/3 * 0.707 mol = 0.471 mol
m(NaN3) = 0.471 mol * 65.0099 g mol-1 = 30.6 g
Separately, write a balanced equation for the decomposition of nitroglycerine. Finally, write a balanced equation for the decomposition of lead azide used for detonation. In what ways are the reactions for sodium azide, nitroglycerine and lead azide similar?
Model Answer
4 C3H5(NO3)3 → 6 N2 + O2 + 12 CO2 + 10 H2O
Pb(N3)2 → Pb + 3 N2
In all three reactions, the reactants are solid or liquid with small volume. A large volume of nitrogen gas is produced. Nitroglycerin produces other gases. The nitrogen molecule has a triple bond and is very stable. Thus, the reactions are highly exothermic, so that gases produced expand rapidly.
Write a balanced equation for the reaction of sodium azide with sulfuric acid to form hydrazoic acid (HN3) and sodium sulfate.
Model Answer
2 NaN3 + H2SO4 → 2 HN3 + Na2SO4
How many grams of hydrazoic acid are produced if 60,0 g of sodium azide reacts with 100 cm3 of sulfuric acid (c = 3,00 mol dm–3)?
Model Answer
n(NaN3) = 60.0 g / 65.0099 g mol-1 = 0.923 mol
n(H2SO4) = 3,00 mol dm-3 * 0.100 dm3 = 0.300 mol
m(HN3) = 2 * 0.300 mol * 43.028 g mol-1 = 25,8 g